Question

20 G of a sample of Ba(OH)2 is dissolved in 50 ml of 0.1 molar HCL solution the excess of HCL was neutralized with 0.1 molar NaOH the volume of NaOH used was 20cc calculate the percentage of Ba(OH)2 in the sample given that atomic weight of Barium is 137 u​

Answer 1

Answer:

Excess HCl left after reaction with Ba(OH)

2

=

20 mL 0.2 N NaOH=x mL 0.5 N HCl

V

1

N

1

=V

2

N

2

20×0.2=x×0.5

x=8 mL

Only 10−8=2 mL HCl reacted with the Ba(OH)

2

sample.

2 mL 0.5 N HCl=

1000

2×0.5

=0.001 moles

Now, Ba(OH)

2

+2HCl→BaCl

2

+H

2

O

Ratio is 1 : 2

x : 0.001

x=0.0005 moles

Weight of Ba(OH)

2

in sample

=molar weight×0.0005

=171×0.0005 g=0.0855 g

%Ba(OH)

2

in sample =

weight of the sample

weight of Ba(OH)

2

in the sample

×100

=

20

0.0855

×100

=0.4275 %